Mathematics - Vector Question with Solution | TestHub

$\begin{array}{rl}& \text{ Area of parallelogram }=\frac{1}{2}|{\overrightarrow{\mathrm{d}}}_1\times {\overrightarrow{\mathrm{d}}}_2\\ & \mathrm{A}=\frac{1}{2}|(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\times (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\frac{\sqrt{21}}{2}\\ & \text{ so, }\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+2\hat{\mathrm{k}}\\ & \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+\beta \hat{\mathrm{j}}\\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\times (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& \alpha & 2\\ -1& \beta & 0\end{array}\right|\end{array}$ $\begin{array}{rl}& =\hat{\mathrm{i}}(-2\beta )-\hat{\mathrm{j}}(2)+\hat{\mathrm{k}}(\beta +\alpha )\\ & |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})\times (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\sqrt{4{\beta }^2+4+(\alpha +\beta {)}^2}=\sqrt{21}\\ & 4{\beta }^2+4+{\alpha }^2+{\beta }^2+2\alpha \beta =21\\ & {\alpha }^2+5{\beta }^2-12=17\\ & {\alpha }^2+5{\beta }^2=29\end{array}$ and $\alpha \beta =-6$ and given ${\alpha }_{\mathrm{i}}\beta$ are integers so, $\alpha =-3,\beta =2$ or $\begin{array}{rl}& \alpha =3,\beta =-2\\ & \left({\alpha }_1,{\beta }_1\right)=(-3,2)\\ & \left({\alpha }_2,{\beta }_2\right)=(3,-2)\\ & {\alpha }_1^2+{\beta }_1^2-{\alpha }_2{\beta }_2=9+4+6=19\end{array}$